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I have two types of customers (type 1 and type 2) enter a shop. Their arrival processes are independent and follow Poisson process with the arrival rates of $\lambda_1$ and $\lambda_2.$

Consider two events where $A = \{\text{customer } q+1 \text{ is type 1}\}$ and $B = \{\text{more than } q \text{ customers from all types arrive} \\ \text{ within a specified time}\}.$

What is the probability of $P(A,B)$? I want mathematical formulation for this problem.

Thanks in advance for considering my question.

kjetil b halvorsen
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Ebi1990
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  • It's easy to solve this problem via Monte Carlo simulation. – Digio Feb 08 '19 at 22:45
  • Do you mean the probability that both events $A$ and $B$ occur? – Michael Hardy Feb 10 '19 at 01:20
  • @Digio : This should admit an easy closed-form solution. (Provided we can be sure exactly what question is intended.) – Michael Hardy Feb 10 '19 at 01:21
  • I suspect you actually meant $\Pr(A\mid B),$ the conditional probability of $A$ given $B,$ since otherwise you'd probably have said something about how much time has passed. – Michael Hardy Feb 10 '19 at 01:22
  • Some things are unclear in your question. If customers arrive at rate $\lambda,$ then the average time until the next customer arrives is $1/\lambda.$ The event $B$ seems to be that at least $q$ customers have arrived. Does that mean by some particular time that many have arrived? Or before some particular event happens? Are you asking about the conditional probability $\Pr(A\mid B)$? Or about $\Pr(A\ \&\ B)$? Or something else? That is not clear. However the question about the probability that the $(q+1)$th customer is of type $1$ is clear. – Michael Hardy Feb 10 '19 at 03:02
  • @MichaelHardy This is Pr(A & B). I want a probability that shows more than q customer arrive and customer q+1 is type 1. Let me ask my question this way. There are two arrival process between time 0 to time T. What is probability of having more that q customer and customer q+1 is type 1? – Ebi1990 Feb 11 '19 at 15:52
  • The probability of getting more than $q$ arrivals before some specified time $t$ is $$ \sum_{x\,=\,q+1}^\infty \frac{((\lambda_1 + \lambda_2)t)^x e^{-(\lambda_1+\lambda_2)t}}{x!} = 1 - \sum_{x\,=\,0}^q \frac{((\lambda_1 + \lambda_2)t)^x e^{-(\lambda_1+\lambda_2)t}}{x!}. $$ – Michael Hardy Feb 25 '19 at 00:32
  • At any particular time, the probability that the next arrival is of Type 1 is $\lambda_1/(\lambda_1 + \lambda_2).$ Showing that that event is independent of whether a specified number of arrivals happen by a specified time would cost some words; otherwise I'd probably have posted on it by now. – Michael Hardy Feb 25 '19 at 00:34

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